#include<bits/stdc++.h>
using namespace std;

const int MN = 200010;

int N, Q;
vector<int> B;
vector<vector<int> > query;

struct BIT {
    vector<int> tree;
    void init() {
        tree = vector<int>(4*MN, 1e9);
    }
    void upd(int idx, int val, int l, int r, int n) {
        if(idx < l || r < idx) return;
        if(l == r) {
            tree[n] = min(tree[n], val);
            return;
        }
        int m = (l + r)>>1;
        upd(idx, val, l, m, 2*n);
        upd(idx, val, m + 1, r, 2*n + 1);
        tree[n] = min(tree[2*n], tree[2*n + 1]);
    }
    int quer(int a, int b, int l, int r, int n) {
        if(b < l || r < a) return 2e9;
        if(a <= l && r <= b) return tree[n];
        int m = (l + r)>>1;
        int L = quer(a, b, l, m, 2*n);
        int R = quer(a, b, m + 1, r, 2*n + 1);
        return min(L, R);
    }
} bit;

int main() {
    scanf("%d", &N);

    B.resize(N + 1);
    int sum = 0;
    for(int i = 1; i <= N; i++) {
        scanf("%d", &B[i]);
        sum += B[i] ^ 1;
    }

    scanf("%d", &Q);

    query.resize(N + 1);
    for(int i = 0; i < Q; i++) {
        int l, r; scanf("%d %d", &l, &r);

        query[l].push_back(r);
    }
    for(int i = 1; i <= N; i++) {
        sort(query[i].begin(), query[i].end());
    }

    bit.init();
    bit.upd(0, 0, 0, MN - 1, 1);
    for(int i = 1; i <= N; i++) {
        for(int j = query[i].size(); j--;) {
            int r = query[i][j];

            bit.upd(r, bit.quer(i - 1, r, 0, MN - 1, 1), 0, MN - 1, 1);
        }
        bit.upd(i, (B[i]? 1 : -1) + bit.quer(i - 1, i - 1, 0, MN - 1, 1), 0, MN - 1, 1);
    }
    cout << sum + bit.quer(N, N, 0, MN - 1, 1);
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int MN = 200010;

int N, Q;
vector<int> B;
vector<vector<int> > query;

struct BIT {
    vector<int> tree;
    void init() {
        tree = vector<int>(4*MN, 1e9);
    }
    void upd(int idx, int val, int l, int r, int n) {
        if(idx < l || r < idx) return;
        if(l == r) {
            tree[n] = min(tree[n], val);
            return;
        }
        int m = (l + r)>>1;
        upd(idx, val, l, m, 2*n);
        upd(idx, val, m + 1, r, 2*n + 1);
        tree[n] = min(tree[2*n], tree[2*n + 1]);
    }
    int quer(int a, int b, int l, int r, int n) {
        if(b < l || r < a) return 2e9;
        if(a <= l && r <= b) return tree[n];
        int m = (l + r)>>1;
        int L = quer(a, b, l, m, 2*n);
        int R = quer(a, b, m + 1, r, 2*n + 1);
        return min(L, R);
    }
} bit;

int main() {
    scanf("%d", &N);

    B.resize(N + 1);
    int sum = 0;
    for(int i = 1; i <= N; i++) {
        scanf("%d", &B[i]);
        sum += B[i] ^ 1;
    }

    scanf("%d", &Q);

    query.resize(N + 1);
    for(int i = 0; i < Q; i++) {
        int l, r; scanf("%d %d", &l, &r);

        query[l].push_back(r);
    }
    for(int i = 1; i <= N; i++) {
        sort(query[i].begin(), query[i].end());
    }

    bit.init();
    bit.upd(0, 0, 0, MN - 1, 1);
    for(int i = 1; i <= N; i++) {
        for(int j = query[i].size(); j--;) {
            int r = query[i][j];

            bit.upd(r, bit.quer(i - 1, r, 0, MN - 1, 1), 0, MN - 1, 1);
        }
        bit.upd(i, (B[i]? 1 : -1) + bit.quer(i - 1, i - 1, 0, MN - 1, 1), 0, MN - 1, 1);
    }
    cout << sum + bit.quer(N, N, 0, MN - 1, 1);
}

```