Problem: While discussing a proper problem A for a Codeforces Round, Kostya created a cyclic array of positive integers a_1, a_2, …, a_n. Since the talk was long and not promising, Kostya created a new cyclic array b_1, b_2, …, b_{n} so that b_i = (a_i mod a_{i + 1}), where we take a_{n+1} = a_1. Here mod is the [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation). When the talk became interesting, Kostya completely forgot how array a had looked like. Suddenly, he thought that restoring array a from array b would be an interesting problem (unfortunately, not A).

Input

The first line contains a single integer n (2 ≤ n ≤ 140582) — the length of the array a.

The second line contains n integers b_1, b_2, …, b_{n} (0 ≤ b_i ≤ 187126).

Output

If it is possible to restore some array a of length n so that b_i = a_i mod a_{(i mod n) + 1} holds for all i = 1, 2, …, n, print «YES» in the first line and the integers a_1, a_2, …, a_n in the second line. All a_i should satisfy 1 ≤ a_i ≤ 10^{18}. We can show that if an answer exists, then an answer with such constraint exists as well.

It it impossible to restore any valid a, print «NO» in one line.

You can print each letter in any case (upper or lower).

Examples

Input

4
1 3 1 0


Output

YES
1 3 5 2


Input

2
4 4


Output

NO

Note

In the first example:

  * 1 mod 3 = 1 
  * 3 mod 5 = 3 
  * 5 mod 2 = 1 
  * 2 mod 1 = 0 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * x % mod;
    x = x * x % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
  return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
  if (B == 0) return A;
  return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
  return A * B / gcd(A, B);
}
};  // namespace Mymath
namespace fwt {
using namespace Mymath;
void FWT(int a[], int n, long long mod) {
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
      }
}
void UFWT(int a[], int n, long long mod) {
  long long rev = inv(2, mod);
  for (int d = 1; d < n; d <<= 1)
    for (int m = d << 1, i = 0; i < n; i += m)
      for (int j = 0; j < d; j++) {
        int x = a[i + j], y = a[i + j + d];
        a[i + j] = 1LL * (x + y) * rev % mod,
              a[i + j + d] = (1LL * (x - y) * rev % mod + mod) % mod;
      }
}
void solve(int a[], int b[], int n, long long mod) {
  FWT(a, n, mod);
  FWT(b, n, mod);
  for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * b[i] % mod;
  UFWT(a, n, mod);
}
};  // namespace fwt
namespace Maxflow {
const int Maxn = 1005;
struct edge {
  int to, rev, cap;
  edge(int _to, int _rev, int _cap) { to = _to, rev = _rev, cap = _cap; }
};
int src = 1002, snk = 1003;
vector<edge> G[Maxn];
int dist[Maxn], seen[Maxn];
int n;
void Clear() {
  for (int i = 0; i < Maxn; i++) G[i].clear();
}
void bfs(int s) {
  for (int i = 0; i < Maxn; i++) dist[i] = -1;
  dist[s] = 0;
  queue<int> Q;
  Q.push(s);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < G[x].size(); i++) {
      edge &e = G[x][i];
      if (e.cap && dist[e.to] == -1) {
        dist[e.to] = dist[x] + 1;
        Q.push(e.to);
      }
    }
  }
}
int dfs(int now, int dest, int fl) {
  if (now == dest) return fl;
  for (int &i = seen[now]; i < G[now].size(); i++) {
    edge &ed = G[now][i];
    if (ed.cap && dist[ed.to] == dist[now] + 1) {
      int f = dfs(ed.to, dest, min(fl, ed.cap));
      if (f) {
        ed.cap -= f;
        G[ed.to][ed.rev].cap += f;
        return f;
      }
    }
  }
  return 0;
}
int dinic(int s = src, int t = snk) {
  int ret = 0;
  while (1) {
    memset(seen, 0, sizeof(seen));
    bfs(s);
    if (dist[t] == -1) return ret;
    int f = dfs(s, t, 2147483647);
    while (f) {
      ret += f;
      f = dfs(s, t, 2147483647);
    }
  }
}
};  // namespace Maxflow
namespace Geometry {
struct iP {
  long long x, y;
  iP(long long _x, long long _y) { x = _x, y = _y; }
  iP() { x = y = 0; }
  iP operator+(iP p) { return iP(x + p.x, y + p.y); }
  iP operator-(iP p) { return iP(x - p.x, y - p.y); }
  iP operator*(long long k) { return iP(x * k, y * k); }
  long long det(iP p) { return x * p.y - y * p.x; }
  long long dot(iP p) { return x * p.x + y * p.y; }
  long long dist() { return x * x + y * y; }
  bool operator<(const iP &p) const {
    if (x != p.x) return x < p.x;
    return y < p.y;
  }
};
struct dP {
  double x, y;
  dP(double _x, double _y) { x = _x, y = _y; }
  dP() { x = y = 0.0; }
  dP operator+(dP p) { return dP(x + p.x, y + p.y); }
  dP operator-(dP p) { return dP(x - p.x, y - p.y); }
  dP operator*(double k) { return dP(x * k, y * k); }
  double det(dP p) { return x * p.y - y * p.x; }
  double dot(dP p) { return x * p.x + y * p.y; }
  double dist() { return x * x + y * y; }
  bool operator<(const dP &p) const {
    if (x != p.x) return x < p.x;
    return y < p.y;
  }
};
vector<iP> convex_hull(iP X[], int l) {
  vector<iP> ret;
  sort(X + 1, X + 1 + l);
  ret.push_back(X[1]);
  for (int i = 2; i <= l; i++) {
    while (ret.size() >= 2 &&
           (ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
      ret.pop_back();
    }
    ret.push_back(X[i]);
  }
  int rs = ret.size() + 1;
  for (int i = l - 1; i >= 1; i--) {
    while (ret.size() >= rs &&
           (ret[ret.size() - 2] - X[i]).det(ret[ret.size() - 1] - X[i]) >= 0) {
      ret.pop_back();
    }
    ret.push_back(X[i]);
  }
  return ret;
}
};  // namespace Geometry
const int Maxn = 150005;
int n;
int a[Maxn];
long long b[Maxn];
int main() {
  n = read();
  bool flag = true;
  bool f2 = true;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i]) flag = false;
    if (a[i] != a[1]) f2 = false;
  }
  if (flag) {
    printf("YES\n");
    for (int i = 1; i <= n; i++) {
      printf("%d ", 1);
    }
    puts("");
    return 0;
  }
  if (f2) {
    printf("NO\n");
    return 0;
  }
  int pos;
  for (int i = 1; i <= n; i++) {
    int nxt = i - 1;
    if (nxt < 1) nxt = n;
    if (a[i] > a[nxt]) {
      pos = i;
      break;
    }
  }
  b[pos] = a[pos];
  int pv = pos - 1;
  if (pv < 1) pv = n;
  long long inf = 1e10;
  b[pv] = inf * a[pos] + a[pv];
  for (int i = pv - 1; i != pos; i--) {
    if (i < 1) i = n;
    if (i == pos) break;
    int nx = i + 1;
    if (nx > n) nx = 1;
    b[i] = b[nx] + a[i];
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) {
    printf("%I64d ", b[i]);
  }
  puts("");
  return ~~(0 ^ 0 ^ 0);
}