Problem: You have c_1 letters 'a', c_2 letters 'b', ..., c_{26} letters 'z'. You want to build a beautiful string of length n from them (obviously, you cannot use the i-th letter more than c_i times). Each c_i is greater than n/3.

A string is called beautiful if there are no palindromic contiguous substrings of odd length greater than 1 in it. For example, the string "abacaba" is not beautiful, it has several palindromic substrings of odd length greater than 1 (for example, "aca"). Another example: the string "abcaa" is beautiful.

Calculate the number of different strings you can build, and print the answer modulo 998244353.

Input

The first line contains one integer n (3 ≤ n ≤ 400).

The second line contains 26 integers c_1, c_2, ..., c_{26} (n/3 < c_i ≤ n).

Output

Print one integer — the number of strings you can build, taken modulo 998244353.

Examples

Input


4
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2


Output


422500


Input


3
2 2 2 2 2 2 3 3 3 2 2 2 2 2 2 3 3 3 2 2 3 2 2 3 2 2


Output


16900


Input


400
348 322 247 158 209 134 151 267 268 176 214 379 372 291 388 135 147 304 169 149 193 351 380 368 181 340


Output


287489790

Solution (in CPP):

Solution:
//*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//*/
 
#include <bits/stdc++.h>
 
// #include <atcoder/all>
 
using namespace std;
// using namespace atcoder;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
typedef long long ll;
// #define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(int)(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
void YES(bool t=true) {cout<<(t?"YES":"NO")<<endl;}
void Yes(bool t=true) {cout<<(t?"Yes":"No")<<endl;}
void yes(bool t=true) {cout<<(t?"yes":"no")<<endl;}
void NO(bool t=true) {cout<<(t?"NO":"YES")<<endl;}
void No(bool t=true) {cout<<(t?"No":"Yes")<<endl;}
void no(bool t=true) {cout<<(t?"no":"yes")<<endl;}
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
int popcount(ll t) { return __builtin_popcountll(t); }
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {0, 1, 0, -1}, dy = {-1, 0, 1, 0};
// vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
struct Setup_io {
    Setup_io() {
        ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        cout << fixed << setprecision(25);
    }
} setup_io;
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'

const int MAXN = 555555;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  if (p < 0) {
    x = mod_pow(x, M - 2, M);
    p = -p;
  }
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = 1;
  for (ll i = 1; i <= n; i++) {
    fact[i] = i * fact[i - 1] % M;
  }
  rfact[n] = mod_inverse(fact[n], M);
  for (ll i = n - 1; i >= 0; i--) {
      rfact[i] = (i + 1) * rfact[i + 1] % M;
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  assert(fact[2] == 2);
  ll ret = fact[n];
  if (rfact[r] == 0) {
    rfact[r] = mod_inverse(fact[r], M);
  }
  ret = (ret*rfact[r]) % M;
  if (rfact[n - r] == 0) {
    rfact[n - r] = mod_inverse(fact[n - r], M);
  }
  ret = (ret*rfact[n - r]) % M;
  return ret;
}

ll nHr(ll n, ll r) {
  return nCr(n+r-1, r);
}

ll solve0(int n, vl c) {
    return 26LL * 26 * mod_pow(25, n - 2) % MOD;
}

int dp1[401][2][2][401];
int sp1[401];

ll solve1(int n, vl c) {
    dp1[2][0][0][0] = 25 * 25;
    dp1[2][1][0][1] = 25;
    dp1[2][0][1][1] = 25;
    dp1[2][1][1][2] = 1;

    for (int i = 2; i < n; i++) {
        rep (j1, 2) rep (j2, 2) {
            rep (k, i + 1) {
                if (dp1[i][j1][j2][k] == 0) continue;

                rep (nj2, 2) {
                    if (j1 == 1 and nj2 == 1) continue;
                    ll mul = 1;
                    if (nj2 == 0) {
                        if (j1 == 0) mul = 24;
                        else mul = 25;
                    }
                    int nk = k + nj2;
                    (dp1[i + 1][j2][nj2][nk] += (ll)dp1[i][j1][j2][k] * mul % MOD) %= MOD;
                }
            }
        }
    }

    rep (i, n + 1) {
        rep (j1, 2) rep (j2, 2) {
            (sp1[i] += dp1[n][j1][j2][i]) %= MOD;
        }
    }

    ll res = 0;
    rep (i, 26) {
        int x = c[i];
        for (int j = x + 1; j <= n; j++) {
            res += sp1[j];
        }
    }
    return res % MOD;
}



int dp2[2][3][3][401][401];
int sp2[401][401];

ll solve2(int n, vl c) {
    rep (i, 3) rep (j, 3) {
        int cnt = 0;
        if (i == 0) cnt++;
        if (j == 0) cnt++;
        int c1 = 0, c2 = 0;
        if (i == 1) c1++;
        if (i == 2) c2++;
        if (j == 1) c1++;
        if (j == 2) c2++;
        ll temp = mod_pow(24, cnt);
        dp2[0][i][j][c1][c2] = temp;
    }

    int s = 0, t = 1;
    for (int i = 2; i < n; i++) {
        rep (j1, 3) rep (j2, 3) {
            rep (k1, i + 1) rep (k2, i + 1) {
                if (k1 + k2 > i + 2) break;
                dp2[t][j1][j2][k1][k2] = 0;
            }
        }

        rep (j1, 3) rep (j2, 3) {
            rep (k1, i + 1) rep (k2, i + 1) {
                if (k1 + k2 > i) break;
                if (dp2[s][j1][j2][k1][k2] == 0) continue;

                rep (nj2, 3) {
                    if (j1 == 1 and nj2 == 1) continue;
                    if (j1 == 2 and nj2 == 2) continue;
                    ll mul = 1;
                    if (nj2 == 0) {
                        if (j1 == 0) mul = 23;
                        else mul = 24;
                    }
                    int nk1 = k1, nk2 = k2;
                    if (nj2 == 1) nk1++;
                    if (nj2 == 2) nk2++;

                    (dp2[t][j2][nj2][nk1][nk2] += (ll)dp2[s][j1][j2][k1][k2] * mul % MOD) %= MOD;
                }
            }
        }
        swap(s, t);
    }

    rep (k1, n + 1) rep (k2, n + 1) {
        rep (j1, 3) rep (j2, 3) {
            (sp2[k1][k2] += dp2[s][j1][j2][k1][k2]) %= MOD;
        }
    }

    ll res = 0;
    rep (i, 26) {
        REP (j, i + 1, 26) {
            int x1 = c[i], x2 = c[j];
            for (int k1 = x1 + 1; k1 <= n; k1++) {
                for (int k2 = x2 + 1; k2 <= n; k2++) {
                    res += sp2[k1][k2];
                }
            }
        }
    }
    return res % MOD;
}


signed main() {
    int n;
    cin >> n;
    vl c(26);
    rep (i, 26) cin >> c[i];

    ll ans0 = solve0(n, c);
    DEBUG(ans0);
    ll ans1 = solve1(n, c);
    DEBUG(ans1);
    ll ans2 = solve2(n, c);
    DEBUG(ans2);

    ll ans = ans0 - ans1 + ans2;
    ans = (ans % MOD + MOD) % MOD;
    cout << ans << endl;

}