Problem: One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump. 

Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.

<image> The picture corresponds to the first example.

The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.

Input

The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100. 

Output

Print single integer a — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.

Examples

Input

ABABBBACFEYUKOTT


Output

4

Input

AAA


Output

1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long MOD = 1000000007;
long double EPS = 1e-10;
long double PI = acos(-1.0);
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <typename T>
T mod(T a, T b) {
  return (a < b ? a : a % b);
}
template <typename T>
T mod_neg(T a, T b) {
  return ((a % b) + b) % b;
}
long long mulmod(long long a, long long b, long long m) {
  long long q = (long long)(((long double)a * (long double)b) / (long double)m);
  long long r = a * b - q * m;
  if (r > m) r %= m;
  if (r < 0) r += m;
  return r;
}
template <typename T>
T expo(T e, T n) {
  T x = 1, p = e;
  while (n) {
    if (n & 1) x = x * p;
    p = p * p;
    n >>= 1;
  }
  return x;
}
template <typename T>
T power(T e, T n, T m) {
  T x = 1 % m, p = e;
  while (n) {
    if (n & 1) x = mod(x * p, m);
    p = mod(p * p, m);
    n >>= 1;
  }
  return x;
}
template <typename T>
T extended_euclid(T a, T b, T &x, T &y) {
  T xx = 0, yy = 1;
  y = 0;
  x = 1;
  while (b) {
    T q = a / b, t = b;
    b = a % b;
    a = t;
    t = xx;
    xx = x - q * xx;
    x = t;
    t = yy;
    yy = y - q * yy;
    y = t;
  }
  return a;
}
template <typename T>
T mod_inverse(T a, T n) {
  T x, y;
  T d = extended_euclid(a, n, x, y);
  return (d > 1 ? -1 : mod_neg(x, n));
}
bool Pvisit[1000006];
vector<int> Primes;
bool isPrime(int N) {
  for (int i = 2; i <= sqrt(N); i++)
    if (N % i == 0) return false;
  return true;
}
void mark(int N) {
  for (int i = N * N; i < 1000006; i += N) Pvisit[i] = false;
}
void sieve() {
  memset(Pvisit, true, 1000006);
  Pvisit[1] = false;
  Primes.push_back(2);
  mark(2);
  for (int i = 3; i <= sqrt(1000006); i += 2)
    if (Pvisit[i] == true) Primes.push_back(i), mark(i);
  for (int i = 1001; i < 1000006; i += 2)
    if (isPrime(i)) Primes.push_back(i), Pvisit[i] = true;
}
int countFact(int n, int p) {
  int k = 0;
  while (n >= p) {
    k += n / p;
    n /= p;
  }
  return k;
}
long long pow(int a, int b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long InverseEuler(int n, int MOD) { return pow(n, MOD - 2, MOD); }
long long factMOD(int n, int MOD) {
  long long res = 1;
  while (n > 0) {
    for (int i = 2, m = n % MOD; i <= m; i++) res = (res * i) % MOD;
    if ((n /= MOD) % 2 > 0) res = MOD - res;
  }
  return res;
}
long long C(int n, int r, int MOD) {
  if (countFact(n, MOD) > countFact(r, MOD) + countFact(n - r, MOD)) return 0;
  return (factMOD(n, MOD) * ((InverseEuler(factMOD(r, MOD), MOD) *
                              InverseEuler(factMOD(n - r, MOD), MOD)) %
                             MOD)) %
         MOD;
}
char vo[] = {'A', 'E', 'I', 'O', 'U', 'Y'};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T;
  T = 1;
  while (T--) {
    string s;
    cin >> s;
    int mx = 1;
    int sm = 1;
    for (int i = 0; i < s.size(); i++) {
      bool v = false;
      for (int j = 0; j < 6; j++)
        if (vo[j] == s[i]) v = true;
      if (v)
        mx = max(mx, sm), sm = 1;
      else
        sm++;
    }
    cout << max(mx, sm);
  }
  return 0;
}