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;
using ll = long long;
int const N = 1e5 + 10;
ll v[N];
struct seg_tree {
  ll mx[N << 2], sum[N << 2];
  seg_tree() { memset(mx, -1, sizeof(mx)); }
  void up(int r, int ls, int rs) {
    mx[r] = mx[ls] == -1 ? mx[rs] : (v[mx[ls]] > v[mx[rs]] ? mx[ls] : mx[rs]);
    sum[r] = sum[ls] + sum[rs];
  }
  void readin(int r, int L, int R) {
    if (L > R) return;
    if (L == R) {
      scanf("%I64d", v + L);
      mx[r] = L, sum[r] = v[L];
      return;
    }
    int M = (L + R) >> 1, ls = r << 1, rs = ls ^ 1;
    readin(ls, L, M), readin(rs, M + 1, R);
    up(r, ls, rs);
  }
  int query_max(int r, int L, int R, int a, int b) {
    if (a <= L && R <= b) return mx[r];
    int M = (L + R) >> 1, ls = r << 1, rs = ls ^ 1, r1 = -1, r2 = -1;
    if (M >= a) r1 = query_max(ls, L, M, a, b);
    if (M < b) r2 = query_max(rs, M + 1, R, a, b);
    return r1 == -1 ? r2 : (v[r1] > v[r2] ? r1 : r2);
  }
  ll query_sum(int r, int L, int R, int a, int b) {
    if (a <= L && R <= b) return sum[r];
    int M = (L + R) >> 1, ls = r << 1, rs = ls ^ 1;
    ll ret = 0;
    if (M >= a) ret += query_sum(ls, L, M, a, b);
    if (M < b) ret += query_sum(rs, M + 1, R, a, b);
    return ret;
  }
  void modify(int r, int L, int R, int p, ll x) {
    if (p < L || p > R || L > R) return;
    if (L == R) {
      v[p] = x, mx[r] = p, sum[r] = x;
      return;
    }
    int M = (L + R) >> 1, ls = r << 1, rs = ls ^ 1;
    if (p <= M)
      modify(ls, L, M, p, x);
    else
      modify(rs, M + 1, R, p, x);
    up(r, ls, rs);
  }
} T;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  T.readin(1, 1, n);
  for (int op, l, r, x, p; m--;) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &l, &r);
      printf("%I64d\n", T.query_sum(1, 1, n, l, r));
    } else if (op == 2) {
      scanf("%d%d%d", &l, &r, &x);
      while (true) {
        ll idx = T.query_max(1, 1, n, l, r);
        if (v[idx] < x) break;
        T.modify(1, 1, n, idx, v[idx] % x);
      }
    } else {
      scanf("%d%d", &p, &x);
      T.modify(1, 1, n, p, x);
    }
  }
  return 0;
}