/*
ans[i][j][b]表示防御力为b的怪物
表示用前i种技能，造成恰好j的伤害消耗的最小晶石
ans[i][j][b]=min{ans[i-1][j-x*(p[i]-b)]+x*k[i]}
x表示第i种技能取x个(0<=x*(p[i]-b)<=j)
优化：http://blog.csdn.net/wumuzi520/article/details/7014830
ans[i][j][b]=min(ans[i-1][j][b],ans[i][j-(p[i]-b)][b]+k[i]【j>=p[i]-b】)
=>ans[j][b]=min(ans[j][b],ans[j-(p[i]-b)][b]+k[i] **i从小到大
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
LL n,m;
LL maxb,maxp,bx,maxa;
LL a[100100],b[100100];
LL ans[10001][11];
LL k[1010],p[1010],ans1;
int main()
{
    LL i,j,t,ze=(long long)0;
    while(scanf("%lld%lld",&n,&m)==2)
    {
        maxb=0;maxp=0;maxa=0;
        for(i=1;i<=n;i++)
        {
            scanf("%lld%lld",&a[i],&b[i]);
            maxb=max(maxb,b[i]);
            maxa=max(maxa,a[i]);
        }
        for(i=1;i<=m;i++)
        {
            scanf("%lld%lld",&k[i],&p[i]);
            maxp=max(maxp,p[i]);
        }
        maxa=max(maxa,maxp);
        /*
        去掉以上这句，会让类似
        1 1
        45 10
        4164 327
        的数据得出错误结果
        */
        if(maxb>=maxp)
        {
            printf("-1\n");
            continue;
        }
        memset(ans,0x3f,sizeof(ans));
        memset(ans[0],0,sizeof(ans[0]));
        for(bx=0;bx<=maxb;bx++)
        {
            for(i=1;i<=m;i++)
                for(j=max(ze,p[i]-bx);j<=2*maxa;j++)
            //    {
                    //if(j>=p[i]-bx)
                        ans[j][bx]=min(ans[j][bx],ans[j-p[i]+bx][bx]+k[i]);
            //    }
            t=0x3f3f3f3f;
            for(j=2*maxa;j>=0;j--)
            {
                t=min(t,ans[j][bx]);
                ans[j][bx]=min(ans[j][bx],t);
            }
        }
        ans1=0;
        for(i=1;i<=n;i++)
            ans1+=ans[a[i]][b[i]];
        printf("%lld\n",ans1);


    }
    return 0;
}