#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={0,1,-1,0,1,1,-1,-1,0};
const ll dx[9]={1,0,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


ll mod=1000000007;//一次元ベクトルの高速化dp
vl matmul(vl &dp, vvl &mt){
    ll m=dp.size();
    vl ret(m,0); 
    rep(i,m)rep(j,m){
        ret[i]+=mt[i][j]*dp[j];
        ret[i]%=mod;
    }
    return ret;
}
vvl update(vvl &mt){
    ll m=mt.size();
    vvl ret(m,vl(m,0));
    rep(i,m)rep(j,m)rep(k,m){
        ret[i][j]+=mt[i][k]*mt[k][j];
        ret[i][j]%=mod;
    }
    return ret;
}
void matpow(vl &dp, vvl &mt, ll k){
    ll m=dp.size();
    while(k){
        if(k&1)dp=matmul(dp,mt);
        mt=update(mt);
        k/=2;
    }
}



void solve2(ll n,ll k,vl v){
    v.push_back(0);
    vvl mat(n+1,vl(n+1));
    rep(i,n)mat[0][i]=1;
    rep(i,n-1)mat[i+1][i]=1;
    mat[n][n]=1;mat[n][n-1]=1;
    matpow(v,mat,k-1);
    cout << v[n-1]<<  " " << (v[n-1]+v[n])%MOD <<endl;
}

int main(){
    ll n,k;cin >> n >>k;
    vl v(n);rep(i,n)cin >> v[i];
    if(n<=30){
        rev(all(v));
        solve2(n,k,v);
    }
    else{
        vl dp(k+100);
        rep(i,n)dp[i]=v[i];
        ll now=accumulate(all(v),0LL);
        repl(i,n,k+10){
            dp[i]=now;
            now+=dp[i];
            now-=dp[i-n];
            now=(now+MOD)%MOD;
        }
        ll cum=0;
        rep(i,k)cum+=dp[i],cum%=MOD;
        cout << dp[k-1] <<" " << cum <<endl;
    }
    
}