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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
const int inf = 1e9 + 143;
const long long longinf = 1e18 + 143;
inline int read() {
  int x;
  scanf(" %d", &x);
  return x;
}
const int N = 1 << 17;
long long s[N << 1];
pair<int, int> t[N << 1];
void tset(int x, int v) {
  t[x + N - 1] = pair<int, int>(v, x);
  s[x += N - 1] = v;
  while (x > 1) {
    x >>= 1;
    s[x] = s[x + x] + s[x + x + 1];
    t[x] = max(t[x + x], t[x + x + 1]);
  }
}
pair<int, int> tmax(int l, int r) {
  pair<int, int> res(0, 0);
  for (l += N - 1, r += N - 1; l <= r; l >>= 1, r >>= 1) {
    if (l & 1) umax(res, t[l++]);
    if (~r & 1) umax(res, t[r--]);
  }
  return res;
}
long long tsum(int l, int r) {
  long long res = 0;
  for (l += N - 1, r += N - 1; l <= r; l >>= 1, r >>= 1) {
    if (l & 1) res += s[l++];
    if (~r & 1) res += s[r--];
  }
  return res;
}
int main() {
  int n = read();
  int m = read();
  for (int i = 1; i <= n; i++) {
    int v = read();
    tset(i, v);
  }
  for (int i = 1; i <= m; i++) {
    int t = read();
    if (t == 1) {
      int l = read(), r = read();
      printf("%I64d\n", tsum(l, r));
    } else if (t == 2) {
      int l = read(), r = read(), x = read();
      for (;;) {
        pair<int, int> mx = tmax(l, r);
        if (mx.first < x) break;
        tset(mx.second, mx.first % x);
      }
    } else {
      int k = read(), x = read();
      tset(k, x);
    }
  }
  return 0;
}