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;
const int N = 100010;
set<pair<int, int> > st[22];
set<pair<int, int> >::iterator a, b, c;
int n, m;
long long ans;
long long f(int first) { return 1LL * first * (first + 1) / 2; }
pair<int, int> unite(set<pair<int, int> >::iterator a,
                     set<pair<int, int> >::iterator b, int j) {
  if (b->second >= b->first) ans -= (1LL << j) * f(b->second - b->first + 1);
  if (a->second >= a->first) ans -= (1LL << j) * f(a->second - a->first + 1);
  pair<int, int> p =
      make_pair(min(a->first, b->first), max(a->second, b->second));
  st[j].insert(p);
  ans += (1LL << j) * f(p.second - p.first + 1);
  st[j].erase(*a);
  st[j].erase(*b);
  return p;
}
void print(int j) {
  for (c = st[j].begin(); c != st[j].end(); ++c)
    cerr << '(' << c->first << ' ' << c->second << ") ";
  cerr << '\n';
}
void insert(int j, int i) {
  st[j].insert(make_pair(i, i));
  ans += (1LL << j);
  a = st[j].lower_bound(make_pair(i, i));
  b = a;
  b--;
  pair<int, int> p = make_pair(i, i);
  if (b != st[j].end()) {
    if (b->second == (a->first - 1)) {
      p = unite(a, b, j);
    }
  }
  a = st[j].find(p);
  b = a;
  b++;
  if (b != st[j].end()) {
    if (b->first == (a->second + 1)) p = unite(a, b, j);
  }
}
void erase(int j, int i) {
  if (st[j].empty()) return;
  a = st[j].upper_bound(make_pair(i, i));
  if (a->first <= i && i <= a->second)
    ;
  else
    --a;
  if (a == st[j].end()) {
    return;
  }
  if (a->first <= i && i <= a->second) {
    ans -= (1LL << j) * f(a->second - a->first + 1);
    pair<int, int> p1 = make_pair(a->first, i - 1);
    pair<int, int> p2 = make_pair(i + 1, a->second);
    if (p1.first <= p1.second) {
      st[j].insert(p1);
      ans += (1LL << j) * f(p1.second - p1.first + 1);
    }
    if (p2.first <= p2.second) {
      st[j].insert(p2);
      ans += (1LL << j) * f(p2.second - p2.first + 1);
    }
    st[j].erase(*a);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1, a; i <= n; ++i) {
    cin >> a;
    for (int j = 0; j <= 20; ++j)
      if ((1LL << j) & a) {
        erase(j, i);
        insert(j, i);
      } else
        erase(j, i);
  }
  for (int i = 1, a, b; i <= m; ++i) {
    cin >> b >> a;
    for (int j = 0; j <= 20; ++j)
      if ((1LL << j) & a) {
        erase(j, b);
        insert(j, b);
      } else {
        erase(j, b);
      }
    cout << ans << '\n';
  }
}