[BUG]Spring RedisTemplate使用GenericFastJsonRedisSerializer后，对于put后的值，不能increment

### 问题描述

把GenericFastJsonRedisSerializer配置到RedisTemplate之后，如果put一个Long类型的值作为初始值，比如: 1L，随后执行increment，会报：ERR value is not an integer or out of range

### 期待的正确结果

使用GenericFastJsonRedisSerializer后，可以用Long去初始化redis的value，不会影响后继的increment操作。

#### 附加信息

出错后，查看redis发现其值为字符串“1L”，而不是“1”，redis无法对“1L”进行处理。和jackson对比:

```java
public static void main(String[] args) throws JsonProcessingException {
Long n = 1L;
// fastjson2
System.out.println(JSON.toJSONString(n, JSONWriter.Feature.WriteClassName));
// jackson
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.setVisibility(PropertyAccessor.ALL, JsonAutoDetect.Visibility.ANY);
objectMapper.activateDefaultTyping(LaissezFaireSubTypeValidator.instance, ObjectMapper.DefaultTyping.NON_FINAL);
System.out.println(objectMapper.writeValueAsString(n));
}
===========
// output:
1L
1
```

Start by locating the GenericFastJsonRedisSerializer integration used with Spring RedisTemplate and reproduce the Long put-then-increment case against Redis. Compare the stored value with the expected Redis integer representation and verify that initializing with 1L no longer prevents increment; the issue provides no file or test path.

Written by the indexing model from the issue text.