Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long s[35];
set<int> off_pos[35];
int v[100005];
long long x[100005];
long long parejas(long long n) { return n * (n + 1) / 2; }
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < 30; i++) {
    off_pos[i].insert(0);
    s[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &v[i]);
    if (i) {
      for (int j = 0; j < 30; j++) {
        if (v[i] & (1 << j)) continue;
        int last = *off_pos[j].rbegin();
        s[j] += 1LL * parejas(i - last - 1) * (1 << j);
        off_pos[j].insert(i);
      }
    }
  }
  for (int i = 0; i < 30; i++) {
    int nxt = n + 1;
    int prev = *off_pos[i].rbegin();
    s[i] += 1LL * parejas(nxt - prev - 1) * (1 << i);
  }
  for (int i = 0; i < 30; i++) off_pos[i].insert(n + 1);
  while (q--) {
    int pos, value;
    cin >> pos >> value;
    int prev = v[pos];
    for (int i = 0; i < 17; i++) {
      int n_bit = value & (1 << i);
      int p_bit = prev & (1 << i);
      if (n_bit == p_bit) continue;
      set<int>::iterator it = off_pos[i].lower_bound(pos);
      if (n_bit) {
        int pivot = *it;
        it--;
        int before = *it;
        it++;
        it++;
        int next = *it;
        s[i] -= 1LL * parejas(pivot - before - 1) * (1 << i);
        s[i] -= 1LL * parejas(next - pivot - 1) * (1 << i);
        s[i] += 1LL * parejas(next - before - 1) * (1 << i);
        off_pos[i].erase(pos);
      } else {
        int next = *it;
        it--;
        int before = *it;
        s[i] -= 1LL * parejas(next - before - 1) * (1 << i);
        s[i] += 1LL * parejas(pos - before - 1) * (1 << i) +
                1LL * parejas(next - pos - 1) * (1 << i);
        off_pos[i].insert(pos);
      }
    }
    v[pos] = value;
    long long ans = 0;
    for (int i = 0; i < 17; i++) {
      if (!s[i]) continue;
      ans += s[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}