Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3
1 0 1
1
1 3

Sample Output 1

1

If you choose to perform the operation, a will become \{1, 1, 1\}, for a hamming distance of 1.

Sample Input 2

3
1 0 1
2
1 1
3 3

Sample Output 2

0

If both operations are performed, a will become \{1, 0, 1\}, for a hamming distance of 0.

Sample Input 3

3
1 0 1
2
1 1
2 3

Sample Output 3

1

Sample Input 4

5
0 1 0 1 0
1
1 5

Sample Output 4

2

It may be optimal to perform no operation.

Sample Input 5

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

Sample Output 5

3

Sample Input 6

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

Sample Output 6

5

Sample Input 7

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

Sample Output 7

1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

#define MIN(a,b) ((a)<(b)?(a):(b))

typedef struct range{
  int l,r;
} range;

int cmp(const void *a,const void *b){
  return ((range *)a)->r-((range *)b)->r;
}

void run(void){
  int n;
  scanf("%d",&n);
  int *b=(int *)calloc(n+1,sizeof(int));
  int i;
  for(i=1;i<=n;i++){
    scanf("%d",b+i);
    b[i]+=b[i-1];
  }
  int q;
  scanf("%d",&q);
  range *p=(range *)calloc(q,sizeof(range));
  for(i=0;i<q;i++) scanf("%d%d",&p[i].l,&p[i].r);
  qsort(p,q,sizeof(range),cmp);
  int *dp=(int *)calloc(n+1,sizeof(int));
  int j=0;
  for(i=1;i<=n;i++){
    int local=dp[i-1]+b[i]-b[i-1];
    while(j<q && p[j].r==i){
      int l=p[j].l;
      local=MIN(local,dp[l-1]+(i-l+1)-(b[i]-b[l-1]));
      j++;
    }
    dp[i]=local;
  }
  printf("%d\n",dp[n]);
}

int main(void){
  run();
  return 0;
}