使用操纵符获取Google Image src

Question

我在puppeteer中有一个小脚本，该脚本必须获取出现在google图片中的第一张图片并显示其src，问题是它仅获取预览图片的src而不是完整图片，我该如何获取完整图片的src？

const puppeteer = require('puppeteer');
const fs = require('fs')

let src = []
async function datos() {
    const browser = await puppeteer.launch({
         headless: false,
     });
     const page = await browser.newPage();
     await page.setViewport({
         width:1920,
         height:1080,
         isMobile: false
     })
     await page.goto('https://www.google.com/imghp');
     await page.waitForXPath('/html/body/div[2]/div[2]/div[2]/form/div[2]/div[1]/div[1]/div/div[2]/input')
     await page.click('#sbtc > div > div.a4bIc > input')
     await page.type('#sbtc > div > div.a4bIc > input', 'Tiras Gluco.Freestyle Optium 50 Unidades')
     await page.click('#sbtc > button > div > span > svg')
     await page.waitForXPath('/html/body/div[2]/c-wiz/div[3]/div[1]/div/div/div/div/div[1]/div[1]/div[1]/a[1]/div[1]/img')
     await page.click('#islrg > div.islrc > div:nth-child(3) > a.wXeWr.islib.nfEiy.mM5pbd > div.bRMDJf.islir > img')
     await page.waitForXPath('/html/body/div[2]/c-wiz/div[3]/div[2]/div[3]/div/div/div[3]/div[2]/c-wiz/div[1]/div[1]/div/div[2]/a/img')
     const data = await page.evaluate(() => {
         src = document.querySelector('#Sva75c > div > div > div.pxAole > div.tvh9oe.BIB1wf > c-wiz > div.OUZ5W > div.zjoqD > div > div.v4dQwb > a > img').src
         return src
     })
     console.log(data)
};   
datos()