$$ \begin{gathered} \textbf{gcd}(x_1, x_2) = \textbf{gcd}(x_1, x_2 - x_1) \\\ \textbf{gcd}(x_l, x_{l+1}, x_{l+2} \cdots x_n) = \textbf{gcd}(x_l, x_{l+1}-x_l, x_{l+2}-x_{l+1} \cdots ) \end{gathered} $$

• 算法设计
  $$ \begin{gathered} \textbf{gcd}(\cdots) \quad \text{线段树维护} \\\ [x_l\cdots x_r] + d \quad \text{树状数组维护} \end{gathered} $$

• 初始化，令 $b_i = a_i - a_{i-1}$
  1.1 对差分序列 $b[1\cdots n]$ 建立线段树维护 $\textbf{gcd}$
  1.2 同时初始化树状数组 $\textbf{add}(tr[\cdots], i, b_i)$

• 对于命令 $C \ l \ r \ d$ 用线段树和树状数组同步维护
  2.1 树状数组 $\textbf{add}(l, d), \ \textbf{add}(r+1, -d)$
  2.2 线段树 $\textbf{change}(1, l, d), \ \textbf{change}(1, r+1, -d)$

• 对于命令 $Q \ l \ r$
  $a_l = \textbf{sum}(l), res \leftarrow \textbf{query}(1, l+1, r)$
  $\textbf{gcd}(a_l, res)$

• 坑点，执行 $[l, r] + d$ 的时候，边界条件是 $b_{n+1} - d$
  建线段树的时候，是根据区间 $[1, n+1]$ 来建立线段树

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <bitset>
#include <assert.h>
#include <unordered_map>

using namespace std;
typedef long long ll;

#define Cmp(a, b) memcmp(a, b, sizeof(b))
#define Cpy(a, b) memcpy(a, b, sizeof(b))
#define Set(a, v) memset(a, v, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define _forS(i, l, r) for(set<int>::iterator i = (l); i != (r); i++)
#define _rep(i, l, r) for(int i = (l); i <= (r); i++)
#define _for(i, l, r) for(int i = (l); i < (r); i++)
#define _forDown(i, l, r) for(int i = (l); i >= r; i--)
#define debug_(ch, i) printf(#ch"[%d]: %d\n", i, ch[i])
#define debug_m(mp, p) printf(#mp"[%d]: %d\n", p->first, p->second)
#define debugS(str) cout << "dbg: " << str << endl;
#define debugArr(arr, x, y) _for(i, 0, x) { _for(j, 0, y) printf("%c", arr[i][j]); printf("\n"); }
#define _forPlus(i, l, d, r) for(int i = (l); i + d < (r); i++)
#define lowbit(i) (i & (-i))
#define MPR(a, b) make_pair(a, b)

pair<int, int> crack(int n) {
    int st = sqrt(n);
    int fac = n / st;
    while (n % st) {
        st += 1;
        fac = n / st;
    }

    return make_pair(st, fac);
}

inline ll qpow(ll a, int n) {
    ll ans = 1;
    for(; n; n >>= 1) {
        if(n & 1) ans *= 1ll * a;
        a *= a;
    }
    return ans;
}

template <class T>
inline bool chmax(T& a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
ll ksc(ll a, ll b, ll mod) {
    ll ans = 0;
    for(; b; b >>= 1) {
        if (b & 1) ans = (ans + a) % mod;
        a = (a * 2) % mod;
    }
    return ans;
}

ll ksm(ll a, ll b, ll mod) {
    ll ans = 1 % mod;
    a %= mod;

    for(; b; b >>= 1) {
        if (b & 1) ans = ksc(ans, a, mod);
        a = ksc(a, a, mod);
    }

    return ans;
}

template <class T>
inline bool chmin(T& a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool lexSmaller(vector<T> a, vector<T> b) {
    int n = a.size(), m = b.size();
    int i;
    for(i = 0; i < n && i < m; i++) {
        if (a[i] < b[i]) return true;
        else if (b[i] < a[i]) return false;
    }
    return (i == n && i < m);
}

// ============================================================== //

const int maxn = 500000 + 10;
ll a[maxn], b[maxn];
int n, m;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll tr[maxn];
void add(int x, ll d) {
    for (int i = x; i <= n; i += lowbit(i)) tr[i] += d;
}
ll sum(int x) {
    ll res = 0;
    for (int i = x; i; i -= lowbit(i)) res += tr[i];
    return res;
}

class A {
public:
    int l, r;
    ll val;
} t[maxn<<2];

void build(int p, int l, int r) {
    t[p].l = l; t[p].r = r;
    if (l == r) {
        t[p].val = b[l];
        return;
    }
    int mid = (l+r) >> 1;
    if (l <= mid) build(p<<1, l, mid);
    if (r > mid) build(p<<1 | 1, mid + 1, r);

    t[p].val = gcd(t[p<<1].val, t[p<<1 | 1].val);
}


void change(int p, int x, ll d) {
    if (x == t[p].l && t[p].r == x) {
        t[p].val += d;
        return;
    }
    int mid = (t[p].l + t[p].r) >> 1;
    if (x <= mid) change(p<<1, x, d);
    if (x > mid) change(p<<1 | 1, x, d);
    t[p].val = gcd(t[p<<1].val, t[p<<1 | 1].val);
}

ll query(int p, const int l, const int r) {
    if (l > r) return 0;
    if (l <= t[p].l && t[p].r <= r) return t[p].val;

    int mid = (t[p].l + t[p].r) >> 1;
    ll res = 0;
    if (l <= mid) res = gcd(res, query(p<<1, l, r));
    if (r > mid) res = gcd(res, query(p<<1 | 1, l, r));
    return res;
}

void prework() {
    for (int i = 1; i <= n; i++) {
        add(i, b[i]);
    }
    build(1, 1, n+1);
}

int main() {
    freopen("input.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 1; i <= n; i++) b[i] = a[i] - a[i-1];

    // use tr[...] and seg-tree
    prework();

    // solve command
    while (m--) {
        char cmd[2];
        scanf("%s", cmd);
        if (cmd[0] == 'Q') {
            int l, r;
            scanf("%d%d", &l, &r);
            ll res = query(1, l+1, r);
            ll al = sum(l);
            res = gcd(al, res);
            printf("%lld\n", abs(res));
        }
        else {
            int l, r;
            ll d;
            scanf("%d%d%lld", &l, &r, &d);
            change(1, l, d); change(1, r+1, -d);
            add(l, d); add(r+1, -d);
        }
    }
}