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 long long int maxn = 1e5 + 5;
template <class T>
inline void read(T& x) {
  bool f;
  char ch = getchar();
  for (f = false; !isdigit(ch); ch = getchar())
    if (ch == '-') f = true;
  for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar())
    ;
  x *= f == 1 ? -1 : 1;
}
long long int n, m, a[maxn], tree[maxn << 2], maxx[maxn << 2];
inline void pushup(long long int num) {
  tree[num] = tree[num << 1] + tree[num << 1 | 1];
  maxx[num] = max(maxx[num << 1], maxx[num << 1 | 1]);
}
inline void built(long long int l, long long int r, long long int num) {
  if (l == r) {
    tree[num] = maxx[num] = a[l];
    return;
  }
  long long int mid = (l + r) >> 1;
  built(l, mid, num << 1);
  built(mid + 1, r, num << 1 | 1);
  pushup(num);
}
inline long long int summation(long long int l, long long int r,
                               long long int x, long long int y,
                               long long int num) {
  if (l >= x && r <= y) {
    return tree[num];
  }
  long long int mid = (l + r) >> 1, sum = 0;
  if (x <= mid) {
    sum += summation(l, mid, x, y, num << 1);
  }
  if (mid < y) {
    sum += summation(mid + 1, r, x, y, num << 1 | 1);
  }
  return sum;
}
inline void slove2(long long int l, long long int r, long long int x,
                   long long int y, long long int k, long long int num) {
  if (maxx[num] < k) {
    return;
  }
  if (l == r && l >= x && r <= y) {
    tree[num] %= k;
    maxx[num] %= k;
    return;
  }
  long long int mid = (l + r) >> 1;
  if (x <= mid) {
    slove2(l, mid, x, y, k, num << 1);
  }
  if (mid < y) {
    slove2(mid + 1, r, x, y, k, num << 1 | 1);
  }
  pushup(num);
}
inline void slove3(long long int l, long long int r, long long int x,
                   long long int y, long long int num) {
  if (l == r && l >= x && r <= x) {
    tree[num] = y;
    maxx[num] = y;
    return;
  }
  long long int mid = (l + r) >> 1;
  if (x <= mid) {
    slove3(l, mid, x, y, num << 1);
  }
  if (mid < x) {
    slove3(mid + 1, r, x, y, num << 1 | 1);
  }
  pushup(num);
}
signed main() {
  read(n);
  read(m);
  for (register long long int i = 1; i <= n; i = -~i) {
    read(a[i]);
  }
  long long int opo, x, y, k;
  built(1, n, 1);
  while (m--) {
    read(opo);
    if (opo == 1) {
      read(x);
      read(y);
      printf("%lld\n", summation(1, n, x, y, 1));
    }
    if (opo == 2) {
      read(x);
      read(y);
      read(k);
      slove2(1, n, x, y, k, 1);
    }
    if (opo == 3) {
      read(x);
      read(y);
      slove3(1, n, x, y, 1);
    }
  }
  return 0;
}