#include using namespace std; typedef signed long long ll; #undef _P #define _P(...) (void)printf(__VA_ARGS__) #define FOR(x,to) for(x=0;x<(to);x++) #define FORR(x,arr) for(auto& x:arr) #define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++) #define ALL(a) (a.begin()),(a.end()) #define ZERO(a) memset(a,0,sizeof(a)) #define MINUS(a) memset(a,0xff,sizeof(a)) //------------------------------------------------------- int N,K; int A[151515]; ll dp[15151][3]; ll mo[3]={1000000007,1000000009,1000000021}; int Q; void solve() { int i,j,k,l,r,x,y; string s; cin>>N>>K; FOR(j,3) dp[0][j]=1; FOR(i,N) { cin>>A[i]; if(A[i]) { for(x=15000-A[i];x>=0;x--) { FOR(j,3) { dp[x+A[i]][j]+=dp[x][j]; if(dp[x+A[i]][j]>=mo[j]) dp[x+A[i]][j]-=mo[j]; } } } } cin>>Q; while(Q--) { cin>>i>>y; i--; if(A[i]) { FOR(x,15000-A[i]+1) { FOR(j,3) { dp[x+A[i]][j]+=mo[j]-dp[x][j]; if(dp[x+A[i]][j]>=mo[j]) dp[x+A[i]][j]-=mo[j]; } } } A[i]=y; if(A[i]) { for(x=15000-A[i];x>=0;x--) { FOR(j,3) { dp[x+A[i]][j]+=dp[x][j]; if(dp[x+A[i]][j]>=mo[j]) dp[x+A[i]][j]-=mo[j]; } } } if(dp[K][0]||dp[K][1]||dp[K][2]) cout<<1<