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;
void dv(vector<long long int> g1) {
  for (auto i = g1.begin(); i != g1.end(); ++i) {
    cout << *i << " ";
  }
}
int invert(int a) {
  if (a == 0) {
    return 1;
  } else if (a == 1) {
    return 0;
  } else {
    return -1;
  }
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 2; i <= n; i++) res = res * i % 1000000007;
  return res;
}
long long int nCr(long long int n, long long int r) {
  return (fact(n) / (fact(r) * fact(n - r))) % 1000000007;
}
long long int nCr1(long long int a, long long int b) {
  if (b > a) {
    return 0;
  } else {
    long long int ans = 1;
    for (long long int i = b + 1; i < a + 1; ++i) {
      ans = ans * i % 1000000007;
    }
    return ans;
  }
}
long long int func(long long int a) {
  if (a % 2 == 0) {
    return -1;
  } else {
    return 1;
  }
}
struct segment_tree {
  long long int size;
  vector<pair<long long int, long long int>> maxs;
  void init(long long int n) {
    size = 1;
    while (size < n) size <<= 1;
    for (long long int i = 0; i < 2 * size; i++) {
      maxs.push_back({-1e18, i});
    }
  }
  void set(long long int i, long long int v, long long int x, long long int lx,
           long long int rx) {
    if (rx - lx == 1) {
      maxs[x] = {v, lx};
      return;
    }
    long long int m = (lx + rx) / 2;
    if (i < m) {
      set(i, v, 2 * x + 1, lx, m);
    } else {
      set(i, v, 2 * x + 2, m, rx);
    }
    maxs[x] = (maxs[2 * x + 1].first > maxs[2 * x + 2].first ? maxs[2 * x + 1]
                                                             : maxs[2 * x + 2]);
  }
  void set(long long int i, long long int v) { set(i, v, 0, 0, size); }
  pair<long long int, long long int> maximum(long long int l, long long int r,
                                             long long int x, long long int lx,
                                             long long int rx) {
    if (lx >= r || l >= rx) return {INT_MIN, -1};
    if (lx >= l && rx <= r) {
      return maxs[x];
    }
    long long int m = (lx + rx) / 2;
    auto max_l = maximum(l, r, 2 * x + 1, lx, m);
    auto max_r = maximum(l, r, 2 * x + 2, m, rx);
    return (max_l.first > max_r.first ? max_l : max_r);
  }
  pair<long long int, long long int> maximum(long long int l, long long int r) {
    return maximum(l, r, 0, 0, size);
  }
};
struct segment_tree_sum {
  long long int size;
  vector<long long int> sums;
  void init(long long int n) {
    size = 1;
    while (size < n) size <<= 1;
    sums.assign(2 * size, 0);
  }
  void set(long long int i, long long int v, long long int x, long long int lx,
           long long int rx) {
    if (rx - lx == 1) {
      sums[x] = v;
      return;
    }
    long long int m = (lx + rx) / 2;
    if (i < m) {
      set(i, v, 2 * x + 1, lx, m);
    } else {
      set(i, v, 2 * x + 2, m, rx);
    }
    sums[x] = sums[2 * x + 1] + sums[2 * x + 2];
  }
  void set(long long int i, long long int v) { set(i, v, 0, 0, size); }
  long long int sum(long long int l, long long int r, long long int x,
                    long long int lx, long long int rx) {
    if (lx >= r || l >= rx) return 0;
    if (lx >= l && rx <= r) return sums[x];
    long long int m = (lx + rx) / 2;
    long long int sum_l = sum(l, r, 2 * x + 1, lx, m);
    long long int sum_r = sum(l, r, 2 * x + 2, m, rx);
    return sum_l + sum_r;
  }
  long long int sum(long long int l, long long int r) {
    return sum(l, r, 0, 0, size);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  vector<long long int> a(n);
  for (long long int x = 0; x < n; ++x) {
    cin >> a[x];
  }
  segment_tree max;
  max.init(n);
  segment_tree_sum sum;
  sum.init(n);
  for (long long int i = 0; i < n; ++i) {
    sum.set(i, a[i]);
    max.set(i, a[i]);
  }
  for (long long int i = 0; i < m; ++i) {
    long long int temp;
    cin >> temp;
    if (temp == 1) {
      long long int l, r;
      cin >> l >> r;
      l--;
      cout << sum.sum(l, r) << endl;
    } else if (temp == 3) {
      long long int k, x;
      cin >> k >> x;
      k--;
      sum.set(k, x);
      max.set(k, x);
      a[k] = x;
    } else {
      long long int l, r, x;
      cin >> l >> r >> x;
      l--;
      if (max.maximum(l, r).first >= x) {
        while (max.maximum(l, r).first >= x) {
          long long int te = max.maximum(l, r).second;
          a[te] = a[te] % x;
          max.set(te, a[te]);
          sum.set(te, a[te]);
        }
      }
    }
  }
}