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, v[100000];
set<int> zero[17];
set<int>::iterator it, jt;
int main() {
  scanf("%d%d", &n, &m);
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
    for (int j = 0; j < 17; ++j)
      if (!((v[i] >> j) & 1)) zero[j].insert(i);
  }
  for (int j = 0; j < 17; ++j) {
    zero[j].insert(-1);
    zero[j].insert(n);
    for (it = ++zero[j].begin(); it != zero[j].end(); ++it) {
      jt = it;
      --jt;
      int len = *it - *jt - 1;
      res += (long long)len * (len + 1) / 2 * (1 << j);
    }
  }
  while (m--) {
    int i, val;
    scanf("%d%d", &i, &val);
    --i;
    for (int j = 0; j < 17; ++j) {
      bool A = (v[i] >> j) & 1;
      bool B = (val >> j) & 1;
      if (A == B) continue;
      if (A == 0) {
        jt = it = zero[j].lower_bound(i);
        --jt;
        ++it;
        zero[j].erase(i);
        int lenB = i - *jt - 1;
        int lenA = *it - i;
        res += ((long long)lenB * lenA + lenA) * (1 << j);
      } else {
        jt = it = zero[j].lower_bound(i);
        --jt;
        int len = *it - *jt - 1;
        res -= (long long)len * (len + 1) / 2 * (1 << j);
        len = *it - i - 1;
        res += (long long)len * (len + 1) / 2 * (1 << j);
        len = i - *jt - 1;
        res += (long long)len * (len + 1) / 2 * (1 << j);
        zero[j].insert(i);
      }
    }
    v[i] = val;
    printf("%lld\n", res);
  }
  return 0;
}