#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

#define PB push_back

using namespace std;

const int N = 2e5 + 500;
const int OFF = (1 << 18);
const int INF = 0x3f3f3f3f;

vector < int > v[N];

int q,n,a[N],prf[N],dp[N],tour[3 * OFF];


void update(int i,int a,int b,int lo,int hi,int x){
    if(lo <= a && b <= hi){
        tour[i] = min(tour[i], x);
        return;
    }
    if(a > hi || b < lo) return;
    update(2 * i,a, (a+b) / 2,lo, hi, x);
    update(2 * i + 1,(a+b+2) / 2, b, lo, hi, x);
}

int query(int x){
    x += OFF;
    int ret = INF;
    for(;x;x>>=1)
        ret = min(ret, tour[x]);
    return ret;
}

int main(){
    memset(tour, INF, sizeof(tour));
    scanf("%d",&n);
    for(int i = 0;i<n;i++){
        scanf("%d",a+i);
        prf[i] = !a[i];
        if(i)
            prf[i] += prf[i - 1];
    }
    scanf("%d",&q);
    for(int i = 0;i<q;i++){
        int x,y;scanf("%d%d",&x,&y);
        x--,y--;
        v[y].PB(x);
    }
    for(int i = 0;i<n;i++){
        dp[i] = a[i] + (i ? query(i - 1) + prf[i - 1] : 0);
        sort(v[i].begin(),v[i].end());
        for(int j = 0;j<v[i].size();j++){
            dp[i] = min(prf[i] + (v[i][j] ? query(v[i][j] - 1) : 0), dp[i]);
            if(v[i][j])
                update(1, 0 ,OFF - 1,v[i][j], i - 1, query(v[i][j] - 1));
            else
                update(1, 0 ,OFF - 1,v[i][j], i - 1, 0);
        }
        tour[i + OFF] = min(tour[i + OFF], dp[i] - prf[i]);
    }
    printf("%d\n", query(n - 1) + prf[n - 1]);
}

### Prompt

Your challenge is to write a cpp solution 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 <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

#define PB push_back

using namespace std;

const int N = 2e5 + 500;
const int OFF = (1 << 18);
const int INF = 0x3f3f3f3f;

vector < int > v[N];

int q,n,a[N],prf[N],dp[N],tour[3 * OFF];


void update(int i,int a,int b,int lo,int hi,int x){
    if(lo <= a && b <= hi){
        tour[i] = min(tour[i], x);
        return;
    }
    if(a > hi || b < lo) return;
    update(2 * i,a, (a+b) / 2,lo, hi, x);
    update(2 * i + 1,(a+b+2) / 2, b, lo, hi, x);
}

int query(int x){
    x += OFF;
    int ret = INF;
    for(;x;x>>=1)
        ret = min(ret, tour[x]);
    return ret;
}

int main(){
    memset(tour, INF, sizeof(tour));
    scanf("%d",&n);
    for(int i = 0;i<n;i++){
        scanf("%d",a+i);
        prf[i] = !a[i];
        if(i)
            prf[i] += prf[i - 1];
    }
    scanf("%d",&q);
    for(int i = 0;i<q;i++){
        int x,y;scanf("%d%d",&x,&y);
        x--,y--;
        v[y].PB(x);
    }
    for(int i = 0;i<n;i++){
        dp[i] = a[i] + (i ? query(i - 1) + prf[i - 1] : 0);
        sort(v[i].begin(),v[i].end());
        for(int j = 0;j<v[i].size();j++){
            dp[i] = min(prf[i] + (v[i][j] ? query(v[i][j] - 1) : 0), dp[i]);
            if(v[i][j])
                update(1, 0 ,OFF - 1,v[i][j], i - 1, query(v[i][j] - 1));
            else
                update(1, 0 ,OFF - 1,v[i][j], i - 1, 0);
        }
        tour[i + OFF] = min(tour[i + OFF], dp[i] - prf[i]);
    }
    printf("%d\n", query(n - 1) + prf[n - 1]);
}

```