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

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
using namespace std;

const int MAX_N = 200005;
const int R = 256 * 1024;

vector<int> P[MAX_N];
int n, q, B[MAX_N], T[2 * R];

int get_min(int x, int y) {
    x += R;
    y += R;
    int w = min(T[x], T[y]);
    while(x / 2 != y / 2) {
        if(x % 2 == 0) w = min(w, T[x + 1]);
        if(y % 2 == 1) w = min(w, T[y - 1]);
        x /= 2;
        y /= 2;
    }
    return w;
}

void update(int x, int w) { 
    x += R;
    T[x] = w;
    while(x > 1) {
        x /= 2;
        T[x] = min(T[2 * x], T[2 * x + 1]);
    }
}

int main() {
    for(int i = 1; i < 2 * R; i++) T[i] = MAX_N;
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> B[i];
    cin >> q;
    for(int i = 1; i <= q; i++) {
        int l, r;
        cin >> l >> r;
        P[l].push_back(r);
    }
    int result = 0;
    for(int i = 1; i <= n; i++) if(B[i] == 0) result++;
    update(0, 0);
    for(int i = 1; i <= n; i++) {
        for(int r : P[i]) update(r, get_min(i - 1, r));
        update(i, min(T[i + R], T[i - 1 + R] + (B[i] == 1 ? 1 : -1)));
    }
    cout << result + T[n + R];
    
    return 0;
}