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;
int n, m;
int i, j, k;
struct edge {
  long long max;
  long long sum;
} tree[400010];
int a[100010];
long long query(int x, int l, int r, int L, int R) {
  if (L <= l && R >= r) return tree[x].sum;
  int mid = (l + r) >> 1;
  if (R <= mid)
    return query(x << 1, l, mid, L, R);
  else if (L > mid)
    return query(x << 1 | 1, mid + 1, r, L, R);
  else
    return query(x << 1, l, mid, L, R) + query(x << 1 | 1, mid + 1, r, L, R);
}
void change1(int x, int l, int r, int L, int R, int mod) {
  if (L <= l && R >= r) {
    if (tree[x].max < mod) return;
    if (l == r) {
      tree[x].max = tree[x].sum = tree[x].max %= mod;
      return;
    }
  }
  int mid = (l + r) >> 1;
  if (L <= mid) change1(x << 1, l, mid, L, R, mod);
  if (R > mid) change1(x << 1 | 1, mid + 1, r, L, R, mod);
  tree[x].max = max(tree[x << 1].max, tree[x << 1 | 1].max);
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
}
void change2(int x, int l, int r, int u, int y) {
  if (l == r) {
    tree[x].max = y;
    tree[x].sum = y;
    return;
  }
  int mid = (l + r) >> 1;
  if (u <= mid)
    change2(x << 1, l, mid, u, y);
  else
    change2(x << 1 | 1, mid + 1, r, u, y);
  tree[x].max = max(tree[x << 1].max, tree[x << 1 | 1].max);
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
}
void build(int x, int l, int r) {
  if (l == r) {
    tree[x].max = a[l];
    tree[x].sum = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  tree[x].max = max(tree[x << 1].max, tree[x << 1 | 1].max);
  tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
}
int read() {
  int x = 0, f = 1;
  char c;
  c = getchar();
  if (c == ' ') return x * f;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
int main() {
  cin >> n >> m;
  for (i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  int opt, a, b, c;
  while (m--) {
    opt = read();
    if (opt == 1) {
      a = read();
      b = read();
      cout << query(1, 1, n, a, b) << endl;
    }
    if (opt == 2) {
      a = read();
      b = read();
      c = read();
      change1(1, 1, n, a, b, c);
    }
    if (opt == 3) {
      a = read();
      b = read();
      change2(1, 1, n, a, b);
    }
  }
}