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;
const int maxn = 1e5 + 7;
unsigned long long a[maxn], sum[maxn << 2], mx[maxn << 2];
int n, m;
void build(int p, int l, int r) {
  if (l == r) {
    sum[p] = mx[p] = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  sum[p] = sum[p << 1] + sum[p << 1 | 1];
  mx[p] = max(mx[p << 1], mx[p << 1 | 1]);
}
void modify(int p, int l, int r, int x, unsigned long long v) {
  if (l == r) {
    sum[p] = mx[p] = v;
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    modify(p << 1, l, mid, x, v);
  else
    modify(p << 1 | 1, mid + 1, r, x, v);
  mx[p] = max(mx[p << 1], mx[p << 1 | 1]);
  sum[p] = sum[p << 1] + sum[p << 1 | 1];
}
unsigned long long querysum(int p, int l, int r, int x, int y) {
  if (x <= l && r <= y) return sum[p];
  int mid = l + r >> 1;
  unsigned long long res = 0;
  if (x <= mid) res += querysum(p << 1, l, mid, x, y);
  if (y > mid) res += querysum(p << 1 | 1, mid + 1, r, x, y);
  return res;
}
unsigned long long querymx(int p, int l, int r, int x, int y) {
  if (x <= l && r <= y) return mx[p];
  int mid = l + r >> 1;
  unsigned long long res = 0;
  if (x <= mid) res = max(res, querymx(p << 1, l, mid, x, y));
  if (y > mid) res = max(res, querymx(p << 1 | 1, mid + 1, r, x, y));
  return res;
}
void opmd(int p, int l, int r, int x, int y, unsigned long long md) {
  if (l == r) {
    sum[p] %= md;
    mx[p] %= md;
    return;
  }
  if (mx[p] < md) return;
  int mid = l + r >> 1;
  if (x <= mid) opmd(p << 1, l, mid, x, y, md);
  if (y > mid) opmd(p << 1 | 1, mid + 1, r, x, y, md);
  mx[p] = max(mx[p << 1], mx[p << 1 | 1]);
  sum[p] = sum[p << 1] + sum[p << 1 | 1];
}
unsigned long long rd() {
  unsigned long long s = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s;
}
int main() {
  n = rd();
  m = rd();
  for (int i = 1; i <= n; i++) a[i] = rd();
  build(1, 1, n);
  while (m--) {
    int op = rd(), l, r;
    if (op == 1) {
      l = rd(), r = rd();
      printf("%lld\n", querysum(1, 1, n, l, r));
    } else if (op == 2) {
      unsigned long long x;
      l = rd();
      r = rd();
      x = rd();
      if (x > querymx(1, 1, n, l, r))
        continue;
      else {
        opmd(1, 1, n, l, r, x);
      }
    } else {
      int k = rd();
      unsigned long long x = rd();
      modify(1, 1, n, k, x);
    }
  }
}