Problem:
Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3

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

#ifdef __cplusplus
#include <iostream>
#include <iomanip>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
#else
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#endif

#define rep(i,n) for(int i = 0; i < (n); ++i)
 
#define BIG_PRIME 1000000007
#define BP_remainder(x) ((x) % BIG_PRIME)

#define INF (1 << 30)
#define INF_NEG (INF * -1)
#define INF_LONG (1LL << 62)
#define INF_LONG_NEG (INF_LONG * -1L)

typedef long long ll;
typedef char str255[256];
int StrLen(char *s) { for(int ptr = 0;; ++ptr) if(!s[ptr]) return ptr; } 
// char * の後ろから n 文字削除, 削除後の文字数を返す
int StrDel(char *s, int n) { int l = StrLen(s); s[l - n + 1] = '\0'; return l - n; }

int NumDigits(int i){ char s[20]; return sprintf(s, "%d", i); }
int NumDigitsLL(long long l){ char s[30]; sprintf(s, "%lld", l); return strlen(s); }

int is1stAns = 1;
#ifdef __cplusplus
template <class T> 
void print(T t) { if (is1stAns) cout << t, is1stAns = 0; else cout << " " << t; }
#else
void printI(int    i) { if (is1stAns) printf(  "%d", i), is1stAns = 0; else printf(  " %d", i); }
void printL(ll     l) { if (is1stAns) printf("%lld", l), is1stAns = 0; else printf(" %lld", l); }
void printD(double d) { if (is1stAns) printf("%.9f", d), is1stAns = 0; else printf(" %.9f", d); }
void printC(char   c) { if (is1stAns) printf(  "%c", c), is1stAns = 0; else printf(  " %c", c); }
void printS(char  *s) { if (is1stAns) printf(  "%s", s), is1stAns = 0; else printf(  " %s", s); }
#endif

signed main()
{
#ifdef __cplusplus
    cout << fixed << setprecision(9);
#endif

    int N; scanf("%d", &N);
    int *A = (int *)malloc(sizeof(int) * (N + 9));
    int *B = (int *)malloc(sizeof(int) * (N + 9));

    rep(i, N + 1) scanf("%d", &A[i]);
    rep(i, N) scanf("%d", &B[i]);

    int ans = 0;
/*    rep(i, N){
        if (A[i] >= B[i]){
            ans += B[i];
        } else {
            ans += A[i];
            B[i] -= A[i];
            if (A[i + 1] >= B[i]){
                ans += B[i];
                A[i + 1] -= B[i];
            } else {
                ans += A[i + 1];
                A[i + 1] = 0;
            }
        }
    }
*/

    // 前からだとWAになったので、後ろから倒してみる

    for (int i = N - 1; i >= 0; --i){
        if (A[i + 1] >= B[i]){
            ans += B[i];
        } else {
            ans += A[i + 1]; B[i] -= A[i + 1];
            if (A[i] >= B[i]){
                ans += B[i]; A[i] -= B[i];
            } else {
                ans += A[i]; A[i] = 0;
            }
        }
    }


    printf("%d\n", ans);

    free(A);
    free(B);
    return 0;
}