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;
int a[100005], n, m;
set<int> off[18];
long long ans = 0;
long long sum(int k, int pos) {
  set<int>::iterator lit = off[k].upper_bound(pos);
  set<int>::iterator uit = lit--;
  long long num = (long long)((*uit) - pos);
  num *= (long long)(pos - (*lit));
  num *= (1ll << k);
  return num;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int k = 0; k < 18; ++k) off[k].insert(-1);
  for (int i = 0, x; i < n; ++i) {
    scanf("%d", &x);
    a[i] = x;
    for (int k = 0; k < 18; ++k) {
      if (x & (1 << k)) {
        off[k].insert(i + 1);
        ans += sum(k, i);
        off[k].erase(i + 1);
      } else
        off[k].insert(i);
    }
  }
  for (int k = 0; k < 18; ++k) off[k].insert(n);
  while (m--) {
    int x, v;
    scanf("%d%d", &x, &v);
    --x;
    for (int k = 0; k < 18; ++k) {
      if (((a[x]) & (1 << k)) == (v & (1 << k))) continue;
      if (v & (1 << k)) {
        off[k].erase(x);
        ans += sum(k, x);
      } else {
        ans -= sum(k, x);
        off[k].insert(x);
      }
    }
    a[x] = v;
    printf("%I64d\n", ans);
  }
}