Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct segtree {
  const long long int inf = 1e18;
  int k;
  vector<long long int> sm, mn, mx, lazy;
  segtree(int n) {
    k = 1;
    while (k <= n) k *= 2;
    k *= 2;
    sm.assign(k, 0LL);
    mn.assign(k, 0LL);
    mx.assign(k, 0LL);
    lazy.assign(k, -1LL);
  }
  void assign(int l, int r, int nd, int a, int b, long long int prop,
              long long int x) {
    if (a > r || b < l) {
      if (prop != -1) {
        lazy[nd] = prop;
        mx[nd] = prop;
        mn[nd] = prop;
        sm[nd] = prop * (b - a + 1);
      }
      return;
    }
    if (a >= l && b <= r) {
      lazy[nd] = x;
      mx[nd] = x;
      mn[nd] = x;
      sm[nd] = x * (b - a + 1);
      return;
    }
    int md = (a + b) / 2;
    if (prop == -1) prop = lazy[nd];
    lazy[nd] = -1;
    assign(l, r, 2 * nd, a, md, prop, x);
    assign(l, r, 2 * nd + 1, md + 1, b, prop, x);
    if (2 * nd + 1 < k) {
      sm[nd] = sm[2 * nd] + sm[2 * nd + 1];
      mx[nd] = max(mx[2 * nd], mx[2 * nd + 1]);
      mn[nd] = min(mn[2 * nd], mn[2 * nd + 1]);
    }
  }
  void assign(int l, int r, long long int x) {
    assign(l, r, 1, 0, k / 2 - 1, -1, x);
  }
  void mod(int l, int r, int nd, int a, int b, long long int prop,
           long long int x) {
    if (nd >= k) return;
    if (prop != -1) {
      lazy[nd] = prop;
      mx[nd] = prop;
      mn[nd] = prop;
      sm[nd] = prop * (b - a + 1);
    }
    if (a > r || b < l || mx[nd] < x) {
      return;
    }
    if (a >= l && b <= r && mx[nd] == mn[nd]) {
      long long int y = mx[nd] % x;
      lazy[nd] = y;
      mx[nd] = y;
      mn[nd] = y;
      sm[nd] = y * (b - a + 1);
      return;
    }
    int md = (a + b) / 2;
    if (prop == -1) prop = lazy[nd];
    lazy[nd] = -1;
    mod(l, r, 2 * nd, a, md, prop, x);
    mod(l, r, 2 * nd + 1, md + 1, b, prop, x);
    if (2 * nd + 1 < k) {
      sm[nd] = sm[2 * nd] + sm[2 * nd + 1];
      mx[nd] = max(mx[2 * nd], mx[2 * nd + 1]);
      mn[nd] = min(mn[2 * nd], mn[2 * nd + 1]);
    }
  }
  void mod(int l, int r, long long int x) { mod(l, r, 1, 0, k / 2 - 1, -1, x); }
  long long int sum(int l, int r, int nd, int a, int b, long long int prop) {
    if (nd >= k) return 0LL;
    if (prop != -1) {
      lazy[nd] = prop;
      mx[nd] = prop;
      mn[nd] = prop;
      sm[nd] = prop * (b - a + 1);
    }
    if (a > r || b < l) {
      return 0LL;
    }
    if (a >= l && b <= r) {
      return sm[nd];
    }
    int md = (a + b) / 2;
    if (prop == -1) prop = lazy[nd];
    lazy[nd] = -1;
    long long int rt =
        sum(l, r, 2 * nd, a, md, prop) + sum(l, r, 2 * nd + 1, md + 1, b, prop);
    if (2 * nd + 1 < k) {
      sm[nd] = sm[2 * nd] + sm[2 * nd + 1];
      mx[nd] = max(mx[2 * nd], mx[2 * nd + 1]);
      mn[nd] = min(mn[2 * nd], mn[2 * nd + 1]);
    }
    return rt;
  }
  long long int sum(int l, int r) { return sum(l, r, 1, 0, k / 2 - 1, -1); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  segtree seg(n);
  for (int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    seg.assign(i, i, x);
  }
  while (q--) {
    int tt;
    cin >> tt;
    if (tt == 2) {
      int l, r;
      long long int x;
      cin >> l >> r >> x;
      l--, r--;
      seg.mod(l, r, x);
    } else if (tt == 3) {
      int in;
      long long int x;
      cin >> in >> x;
      in--;
      seg.assign(in, in, x);
    } else {
      int l, r;
      cin >> l >> r;
      l--, r--;
      cout << seg.sum(l, r) << "\n";
    }
  }
}