令dp[i]表示，以第i个点为转折点，i前面全部是大写，后面全部是都是小写的代价，自然等于i前面的小写+i后面的大写。（i自己是大写还是小写无所谓，可以没有代价的归入大写或者小写）
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
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;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
char s[N];
int preup[N], prelow[N], sufup[N], suflow[N];
int main()
{
scanf("%s", s + 1);
int len = strlen(s+1);
upd(i, 1, len)
{
preup[i] = preup[i - 1] + isupper(s[i]);
prelow[i] = prelow[i - 1] + (islower(s[i]) ? 1 : 0);
}
dwd(i, len, 1)
{
sufup[i] = sufup[i + 1] + isupper(s[i]);
suflow[i] = suflow[i + 1] + (islower(s[i]) ? 1 : 0);
}
int ans = INF;
upd(i, 1, len)
{
ans = min(prelow[i - 1] + sufup[i + 1], ans);
}
cout << ans << endl;
return 0;
}