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 = 400010;
long long sm[MAXN], mx[MAXN];
long long N, M, A[100010];
void push_up(long long o) {
  sm[o] = sm[o << 1] + sm[(o << 1) + 1];
  mx[o] = max(mx[o << 1], mx[(o << 1) + 1]);
}
void bul(long long l, long long r, long long o) {
  if (l == r) {
    mx[o] = sm[o] = A[l];
    return;
  }
  long long m = (l + r) >> 1;
  bul(l, m, o << 1);
  bul(m + 1, r, (o << 1) + 1);
  push_up(o);
}
void md(long long ms, long long ll, long long rr, long long l, long long r,
        long long o) {
  if (mx[o] < ms) return;
  if (l == r) {
    sm[o] %= ms;
    mx[o] = sm[o];
    return;
  }
  long long m = (l + r) >> 1;
  if (ll <= m) md(ms, ll, rr, l, m, o << 1);
  if (rr > m) md(ms, ll, rr, m + 1, r, (o << 1) + 1);
  push_up(o);
}
void cng(long long loc, long long k, long long l, long long r, long long o) {
  if (l == r) {
    mx[o] = sm[o] = k;
    return;
  }
  long long m = (l + r) >> 1;
  if (loc <= m)
    cng(loc, k, l, m, o << 1);
  else
    cng(loc, k, m + 1, r, (o << 1) + 1);
  push_up(o);
}
long long qs(long long ll, long long rr, long long l, long long r,
             long long o) {
  if (ll <= l && r <= rr) return sm[o];
  long long m = (l + r) >> 1;
  long long ans = 0;
  if (ll <= m) ans += qs(ll, rr, l, m, o << 1);
  if (rr > m) ans += qs(ll, rr, m + 1, r, (o << 1) + 1);
  return ans;
}
int main() {
  scanf("%lld%lld", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
  bul(1, N, 1);
  for (int i = 1; i <= M; i++) {
    long long a, b, c, d;
    scanf("%lld", &a);
    if (a == 1) {
      scanf("%lld%lld", &b, &c);
      printf("%lld\n", qs(b, c, 1, N, 1));
    } else if (a == 2) {
      scanf("%lld%lld%lld", &b, &c, &d);
      md(d, b, c, 1, N, 1);
    } else if (a == 3) {
      scanf("%lld%lld", &b, &c);
      cng(b, c, 1, N, 1);
    }
  }
  return 0;
}