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>
const long long inf = 1000000000;
const long long inf64 = inf * inf;
const long long base = inf + 7;
const double pi = acos(-1.0);
using namespace std;
int n, m;
vector<long long> a;
struct tree {
  long long sum, mx;
  tree() {
    sum = 0;
    mx = -1;
  }
};
vector<tree> t;
void pushup(long long v) {
  t[v].sum = t[v << 1].sum + t[v << 1 | 1].sum;
  t[v].mx = max(t[v << 1].mx, t[v << 1 | 1].mx);
}
void build(long long v, long long tl, long long tr) {
  if (tl == tr)
    t[v].sum = t[v].mx = a[tl];
  else {
    long long tm = (tl + tr) >> 1;
    build(v << 1, tl, tm);
    build(v << 1 | 1, tm + 1, tr);
    pushup(v);
  }
}
long long sum(long long v, long long tl, long long tr, long long l,
              long long r) {
  if (l <= tl & tr <= r)
    return t[v].sum;
  else {
    long long tm = (tl + tr) >> 1;
    long long L, R;
    L = R = 0;
    if (l <= tm) L = sum(v << 1, tl, tm, l, r);
    if (r > tm) R = sum(v << 1 | 1, tm + 1, tr, l, r);
    pushup(v);
    return (L + R);
  }
}
void up(long long v, long long tl, long long tr, long long l, long long r,
        long long x) {
  if (tl == tr)
    t[v].sum %= x, t[v].mx %= x;
  else {
    long long tm = (tl + tr) >> 1;
    if (l <= tm && t[v << 1].mx >= x) up(v << 1, tl, tm, l, r, x);
    if (r > tm && t[v << 1 | 1].mx >= x) up(v << 1 | 1, tm + 1, tr, l, r, x);
    pushup(v);
  }
}
void put(long long v, long long tl, long long tr, long long pos,
         long long val) {
  if (tl == tr)
    t[v].sum = t[v].mx = val;
  else {
    long long tm = (tl + tr) >> 1;
    if (pos <= tm)
      put(v << 1, tl, tm, pos, val);
    else
      put(v << 1 | 1, tm + 1, tr, pos, val);
    pushup(v);
  }
}
bool solve() {
  scanf("%d %d", &n, &m);
  a.resize(n + 1);
  t.resize(4 * (n + 1), tree());
  for (int i(1); i <= n; i++) scanf("%I64d", &a[i]);
  build(1, 1, n);
  int type, l, r;
  long long x;
  for (int i(0); i < m; i++) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d %d", &l, &r);
      printf("%I64d\n", sum(1, 1, n, l, r));
    } else if (type == 2) {
      scanf("%d %d %I64d", &l, &r, &x);
      up(1, 1, n, l, r, x);
    } else {
      scanf("%d %I64d", &l, &x);
      put(1, 1, n, l, x);
    }
  }
  return true;
}
int main() {
  solve();
  return 0;
}