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;
inline long long read() {
  long long x(0), f(1);
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  int num(0);
  if (x == 0) {
    putchar('0');
    putchar('\n');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  char a[11];
  while (x > 0) {
    a[num] = x % 10 + '0';
    x /= 10;
    ++num;
  }
  for (int i(num - 1); i >= 0; --i) putchar(a[i]);
  putchar('\n');
  return;
}
long long sum[4123456];
long long maxinum[4123456];
inline void pushup(int node) {
  sum[node] = sum[node << 1] + sum[node << 1 | 1];
  maxinum[node] = max(maxinum[node << 1], maxinum[node << 1 | 1]);
}
void build(int l, int r, int node) {
  if (l == r) {
    maxinum[node] = sum[node] = read();
    return;
  }
  int mid((l + r) >> 1);
  build(l, mid, node << 1);
  build(mid + 1, r, node << 1 | 1);
  pushup(node);
  return;
}
void update(int l, int r, int points, long long to, int node) {
  if (l == r) {
    maxinum[node] = sum[node] = to;
    return;
  }
  int mid((l + r) >> 1);
  if (mid >= points) update(l, mid, points, to, node << 1);
  if (mid < points) update(mid + 1, r, points, to, node << 1 | 1);
  pushup(node);
}
void update(int l, int r, int L, int R, int mod, int node) {
  if (maxinum[node] < mod) return;
  if (l == r) {
    sum[node] %= mod;
    maxinum[node] = sum[node];
    return;
  }
  int mid((l + r) >> 1);
  if (mid >= L) update(l, mid, L, R, mod, node << 1);
  if (mid < R) update(mid + 1, r, L, R, mod, node << 1 | 1);
  pushup(node);
}
long long Sum(int l, int r, int L, int R, int node) {
  if (l >= L && r <= R) return sum[node];
  long long _sum(0);
  int mid((l + r) >> 1);
  if (mid >= L) _sum += Sum(l, mid, L, R, node << 1);
  if (mid < R) _sum += Sum(mid + 1, r, L, R, node << 1 | 1);
  return _sum;
}
int main() {
  int n(read()), m(read());
  build(1, n, 1);
  for (int i(1); i <= m; ++i) {
    int cmd(read()), l(read()), r(read());
    if (cmd == 1) {
      write(Sum(1, n, l, r, 1));
    } else if (cmd == 2) {
      int mod(read());
      update(1, n, l, r, mod, 1);
    } else {
      update(1, n, l, r, 1);
    }
  }
  return 0;
}