Problem:
Score : 400 points

Problem Statement

Given is a string S. Each character in S is either a digit (0, ..., 9) or ?.

Among the integers obtained by replacing each occurrence of ? with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.

Since the answer can be enormous, print the count modulo 10^9+7.

Constraints

S is a string consisting of digits (0, ..., 9) and ?.

1 \leq |S| \leq 10^5

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of integers satisfying the condition, modulo 10^9+7.

Sample Input 1

??2??5

Sample Output 1

768

For example, 482305, 002865, and 972665 satisfy the condition.

Sample Input 2

?44

Sample Output 2

1

Only 044 satisfies the condition.

Sample Input 3

7?4

Sample Output 3

0

We may not be able to produce an integer satisfying the condition.

Sample Input 4

?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???

Sample Output 4

153716888

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

using namespace std;

#define MAX             100005
#define MOD             1000000007
#define eps             1e-6
int fx[] =              {1,-1,0,0};
int fy[] =              {0,0,1,-1};

#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0);
#define fRead           freopen("in.txt","r",stdin);
#define fWrite          freopen ("out.txt","w",stdout);

#define ll              long long
#define ull             unsigned long long
#define ff              first
#define ss              second
#define pb              push_back
#define PI              acos(-1.0)
#define mk              make_pair
#define pii             pair<int,int>
#define pll             pair<LL,LL>
#define vi              vector <int>
#define vl              vector <LL>
#define vp              vector<pii>
#define all(a)          a.begin(),a.end()

#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))

#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define IT(it,x)	for(it=x.begin();it!=x.end;it++)

#define MEM(a,x)        memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))
#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())

//#define NUM 1000005bool flag[NUM];vector <int> prime;inline void sieve(){flag[1]=true;for(int i=4; i<NUM; i+=2){flag[i]=true;}for(int i=3; i*i<=NUM; i+=2){if(!flag[i]){for(int j=i*i; j<NUM; j+=2*i){flag[j]=true;}}}prime.pb(2);for(int i=3; i<NUM; i+=2){if(!flag[i]){prime.pb(i);}}return;}

//inline int NOD(ll n){int div=1;for(int i=0; prime[i]*prime[i]<=n; i++){if(n%prime[i]==0){int cnt=0;while(n%prime[i]==0){n/=prime[i];cnt++;}div*=(cnt+1);}}if(n>1){div*=2;}return div;}

//inline ll BigMod(ll B,ll P,ll M){ll R=1;while(P>0){if(P & 1) R=(R*B)%M;P/=2;B=(B*B)%M;}return R%M;}

//inline void build(int L,int R,int pos){if(L==R){tree[pos]=arr[L];return;}int mid=(L+R)/2;build(L,mid,pos*2+1);build(mid+1,R,pos*2+2);tree[pos]=min(tree[pos*2+1],tree[pos*2+2]);return;}

//inline int query(int ql,int qr,int L,int R,int pos){if(ql>R or qr<L) return MAX;else if(ql<=L and qr>=R) return tree[pos];int mid=(L+R)/2;int p=query(ql,qr,L,mid,2*pos+1);int q=query(ql,qr,mid+1,R,2*pos+2);return min(p,q);}

//void updateRange(int b,int e,int L,int R,int pos,ll val){if(lazy[pos]!=0){tree[pos]+=(R-L+1)*lazy[pos];if(L!=R){lazy[2*pos+1]+=lazy[pos];lazy[2*pos+2]+=lazy[pos];}lazy[pos]=0;}if(L>R or L>e or R<b) return;if(L>=b and R<=e){tree[pos]+=(R-L+1)*val;if(L!=R){lazy[2*pos+1]+=val;lazy[2*pos+2]+=val;}return;}int mid=(L+R)/2;updateRange(b,e,L,mid,2*pos+1,val);updateRange(b,e,mid+1,R,2*pos+2,val);tree[pos]=tree[2*pos+1]+tree[2*pos+2];return;}

//ll getSum(int ql,int qr,int L,int R,int pos){if(lazy[pos]!=0){tree[pos]+=(R-L+1)*lazy[pos];if(L!=R){lazy[2*pos+1]+=lazy[pos];lazy[2*pos+2]+=lazy[pos];}lazy[pos]=0;}if(L>R or ql>R or qr<L) return 0;if(L>=ql and qr>=R) return tree[pos];int mid=(L+R)/2;return getSum(ql,qr,L,mid,2*pos+1)+getSum(ql,qr,mid+1,R,2*pos+2);}

//inline void bfs(int s){MEM(vis,0);vis[s]=1;queue<int>Q;Q.push(s);while(!Q.empty()){int x=Q.front();Q.pop();REP(i,V[x].size()){if(!vis[V[x][i]]){vis[V[x][i]]=1;Q.push(V[x][i]);}}}return;}

//int cell[100][100]; int d[100][100],vis[100][100];int row,col;inline void bfs(int sx,int sy){ MEM(vis,0);vis[sx][sy]=1;queue<pii>q; q.push(pii(sx,sy));while(!q.empty()){pii top=q.front();q.pop();for(int k=0; k<4; k++){int tx=top.ff+fx[k];int ty=top.ss+fy[k]; if(tx>=0 and tx<row and ty>=0 and ty<col and cell[tx][ty]!=-1 and vis[tx][ty]==0){vis[tx][ty]=1;d[tx][ty]=d[top.ff][top.ss]+1;q.push(pii(tx,ty)); }}}return;}

//inline void bi_dfs(int s,int c){if(vis[s]) return;vis[s]=1;color[s]=c;REP(i,V[s].size()) dfs(V[s][i],1-c);return;}

//struct point{int name,val;bool operator <(const point &p) const{return p.val < val;}};vector<int>V[105];int dis[105],cost[105][105];priority_queue<point>Q;void Dijkstra(int s){dis[s]=0;point get;get.name=s;get.val=0;Q.push(get);while(!Q.empty()){point tmp=Q.top();Q.pop();int now=tmp.name;REP(i,V[now].size()){int x=V[now][i];if(dis[now]+cost[now][x]<dis[x]){dis[x]=dis[now]+cost[now][x];get.name=x;get.val=dis[x];Q.push(get);}}}return;}

//struct edge{int u,v,w;bool operator < (const edge &p) const{return w < p.w;}};int pr[MAX];vector<edge>e;int find(int r){return pr[r]= (pr[r]==r) ? r:find(pr[r]);}int kruskal(int n){sort(e.begin(),e.end());FOR(i,1,n) pr[i]=i;int cnt=0,sum=0;REP(i,e.size()){int x=find(e[i].u);int y=find(e[i].v);if(x!=y){pr[x]=y;cnt++;sum+=e[i].w;if(cnt==n-1) break;}}return sum;}

//inline ll GCD (ll a,ll b ) {if(a < b) swap(a, b); while ( b ) { a = a % b; swap ( a, b ); } return a;}

//int Set(int N,int pos){return N=N | (1<<pos);}

//int reset(int N,int pos){return N= N & ~(1<<pos);}

//bool check(int N,int pos){return (bool)(N & (1<<pos));}

//// ***************************************************** TEMPLETE ***************************************************** ////

ll N;

ll dp[100005][15];

string str;


ll call(ll pos,ll mod)
{
    if(pos==N)
    {
        if(mod==5)return 1;
        else return 0;
    }
    if(dp[pos][mod]!=-1)return dp[pos][mod];
    ll ret=0;
    if(str[pos]!='?')
    {
        ll num=str[pos]-'0';
        ll d=((mod*10)+num)%13;
        ret=call(pos+1,d)%MOD;
        return dp[pos][mod]=ret;
    }
    REP(i,10)
    {
        ll d=((mod*10)+i)%13;
        ret=(ret+call(pos+1,d))%MOD;
    }
    return dp[pos][mod]=ret;
}

int main()
{
    MEM(dp,-1);
    cin >> str;
    N=str.size();
    ll ans=call(0,0)%MOD;
    cout << ans;
}