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 n, m;
int ai[101000];
set<pair<int, int> > ds[17];
long long ss[17];
inline void ins(int k, int x) {
  pair<int, int> t = make_pair(x, x);
  set<pair<int, int> >::iterator it = ds[k].lower_bound(t);
  int d1 = 0, d2 = 0;
  if (it != ds[k].begin()) {
    set<pair<int, int> >::iterator it2 = it;
    it2--;
    if (it2->second == x - 1) {
      d1 = it2->second - it2->first + 1;
      ds[k].erase(it2);
    }
  }
  if (it != ds[k].end()) {
    if (it->first == x + 1) {
      d2 = it->second - it->first + 1;
      ds[k].erase(it);
    }
  }
  ss[k] += (d1 + 1LL) * (d2 + 1LL);
  ds[k].insert(make_pair(x - d1, x + d2));
}
inline void del(int k, int x) {
  pair<int, int> t = make_pair(x, n + 1);
  set<pair<int, int> >::iterator it = ds[k].lower_bound(t);
  it--;
  ds[k].erase(it);
  pair<int, int> v1 = make_pair(it->first, x - 1);
  pair<int, int> v2 = make_pair(x + 1, it->second);
  ss[k] -= (v1.second - v1.first + 2LL) * (v2.second - v2.first + 2LL);
  if (v1.first <= v1.second) ds[k].insert(v1);
  if (v2.first <= v2.second) ds[k].insert(v2);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (n); i++) {
    int a;
    scanf("%d", &a);
    for (int j = 0; j < (17); j++) {
      if (a & (1 << j)) ins(j, i);
    }
    ai[i] = a;
  }
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    long long ans = 0;
    for (int j = 0; j < (17); j++) {
      if ((ai[x] & (1 << j)) != (y & (1 << j))) {
        if (ai[x] & (1 << j))
          del(j, x);
        else
          ins(j, x);
      }
      ans += ss[j] * (1LL << j);
    }
    ai[x] = y;
    printf("%I64d\n", ans);
  }
}