#include <iostream>
#include <cstdio>
#include <string.h>
#define INF 0x3f3f3f3f
#define MAXN 100000 + 10
#define MAXM 1000 + 10
#define MAXA 1000 + 10
using namespace std;
typedef long long int LL;
int a[MAXN], b[MAXN];
int k[MAXM], p[MAXM];
LL dp[MAXA][20];      //在此防禦力（第二維）下 產生至少為此攻擊（第一維） 最少需要的晶石

int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m;
    int maxA, maxB, maxP;
    while (scanf("%d %d", &n, &m) != EOF)
    {

        maxA = maxB = maxP = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d %d", a + i, b + i);
            maxA = max(maxA, a[i]);
            maxB = max(maxB, b[i]);
        }

        for (int i = 0; i < m; i++)
        {
            scanf("%d %d", k + i, p + i);
            maxP = max(maxP, p[i]);
        }

        if (maxP <= maxB)
        {
            printf("-1\n");
            continue;
        }
        memset(dp, 0x3f, sizeof(dp));
        memset(dp[0], 0, sizeof(dp[0]));

        for (int i = 0; i <= maxB; i++)     //迴圈防禦力
        {
            for (int x = 0; x <= maxA; x++)  //產生的攻擊
            {
                for (int j = 0; j < m; j++)   //考慮j之前的技能
                {
                    int temp = p[j] - i;
                    if (x > temp)
                        dp[x][i] = min(dp[x][i], dp[x - temp][i] + k[j]);
                    else
                        dp[x][i] = min(dp[x][i], (LL)k[j]);
                }

            }

        }
        LL re = 0;
        for (int i = 0; i < n; i++)
            re += dp[a[i]][b[i]];
        printf("%lld\n", re);


    }

}